probability ques

hi,can anyone please help me out with this?thanks in advance!

four letters (each written to a different person) are randomly placed into four envelopes ( these already have the persons' addresses on it).what is the probability that
a) no letter is placed correctly?
b) one letter is placed correctly?

You could use the "brute force" method, and just write out all the possible permutations. Let the envelopes be labelled A, B, C, and D. Ditto for the envelopes.
Keep the envelopes in order and distribute the letters in the order of each possible permutation of the letters. You can just write them out: ABCD (which would mean every letter goes into the right envelope), ABDC (which means the first two are correct), etc. There are 4! = 24 such permutations. Write them all out and you will find that there are 9 in which no letter goes into the correct envelope, so the probability for your question a) is 9/24 or 3/8. There also happen to be 9 permutations in which only one letter goes into the right envelope, so the answer to b) is also 3/8). This is difficult to compute from the rules of probability, because the events are not independent. For example in those events where Letter A goes into envelope B, it becomes impossible for Letter B to go into the correct envelope (it already has letter A in it). Therefore, you reason as follows: There is a 3/4 probability that the first letter will not go into the correct envelope. For the 3/4 of the time this happens, there must be one of the remaining three that cannot possibly go into the correct envelope. For example, if letter A goes into envelope B, there is a 100% chance that letter B will not find its correct envelope. But what about C & D? Well, after letter A is put into envelope B, there are three envelopes left: A, C, and D. No matter which envelope letter B goes into, there will be a 50-50 chance (1/2) that the next-to last letter goes into a wrong envelope. If Letter B goes into envelope A, envelopes C and D will be left, so there is a 50-50 chance that the third one picked will be in the wrong envelope. If that is true, the last letter must also go into the last (wrong envelope), so the overall probability is just 3/4 X 1/2 = 3/8. But suppose that letter B ends up in envelope C? Now envelopes A and D are left for the next choice. It is again a 50-50 chance that letter C goes into envelope D, in which case the D must go into envelope A with 100% probability. Again, the calculation is 3/4 X 1/2 = 3/8.

Oops, I take it back.

Soroban is indeed correct. I was a bit sloppy in writing out the entire sample space. There are only 8 out of 24 permutations in which there are exactly 1 letter in the correct envelope, so 1/3 is the correct answer to the second part of the question.